What is the molar mass of 342 g mol?

Given: Percentage by mass of cane sugar solution = 5 %Percentage by mass of glucose solution = 5 %,Freezing point of cane sugar solution = 271 K

Molar mass of cane sugar = 342 g mol-1

To find: Freezing point of glucose solution

Formula: `"M"_2 = (1000 xx "K"_"f" xx "W"_2)/(triangle "T"_"f" "W"_1)`

Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W2) = 5 g, and mass of solvent (W1) = 95 g.

5 % glucose solution means that mass of glucose = `"W"_2^'` = 5g, and mass of solvent = `"W"_1^'` = 95 g

Molar mass of glucose (C6H12O6) = `("M"_2^')` = 180 g mol-1

Δ Tf for cane sugar solution = `triangle "T"_"f" = "T"_"f"^0 - "T"_"f"` = 273.15 K - 271 K = 215 K

Now, using the formula,

`"M"_2 = (1000 xx "K"_"f" xx "W"_2)/(triangle "T"_"f" "W"_1)`

Rearranging the formula, we get

`1000 "K"_"f" = ("M"_2 xx triangle "T"_"f" xx "W"_1)/"W"_2` ......(1)

`1000 "K"_"f" = ("M"_2^' xx triangle "T"_"f"^' xx "W"_1^')/"W"_2^'` .....(2)

From equations (1) and (2),

`("M"_2 xx triangle "T"_"f" xx "W"_1)/"W"_2 = ("M"_2^' xx triangle "T"_"f"^' xx "W"_1^')/"W"_2^'`

∴ `(342 "mol"^-1 xx 2.15 "K" xx 95 "g")/"5 g" = (180 "g mol"^-1 xx triangle "T"_"f"^' xx 95 "g")/"5 g"`

`triangle "T"_"f"^' = (342 "mol"^-1 xx 2.15 "K")/(180 "g mol"^-1)` = 4.085 K

∴ Freezing point of glucose solution (Tf) = `"T"_"f"^0 - triangle "T"_"f"^'`

= 273.15 K - 4.085 K

= 269.065 K

Freezing point of glucose solution is 269.065 K.

Alternate method:

Formulae: 1. m = `(1000 "W"_2)/("M"_2 "W"_1)`

2. Δ Tf = Kfm

Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W2) = 5 g, and mass of solvent (W1) = 95 g.

Now, using formula (i),

Molality of cane sugar solution = m = `(1000 "W"_2)/("M"_2 "W"_1)`

`= (1000 "g kg"^-1 xx 5 "g")/(342 "g mol"^-1 xx 95 "g")`= 0.1539 m

Now, Δ Tf for cane sugar solution = Δ Tf = `triangle "T"_"f"^0 - "T"_"f"` = 273.15 K - 271 K = 2.15 K

From formula (ii),

Δ Tf (cane sugar) = Kf × m

`"K"_"f" = (triangle "T"_"f")/"m"`

∴ `"K"_"f" = 2.15/0.1539` = 13.97 K kg mol-1

5 % glucose solution means that mass of glucose = `"W"_2^'` = 5 g,

and mass of solvent = `"W"_1^'` = 95 g

Molar mass of glucose (C6H12O6) = `("M"_2^')` = 180 g mol-1

Using formula (i),

Molality of glucose solution =

m = `(1000 "W"_2)/("M"_2 "W"_1)`

= `(1000 "g kg"^-1 xx 5 "g")/(180 "g mol"^-1 xx 95 "g")` = 0.2924 m

From formula (ii),

`triangle "T"_"f"^' ("glucose") = "K"_"f" xx "m"`

(⸪ Since solvent is same, Kf is same for cane sugar and glucose solutions.)

∴ `triangle "T"_"f"^' ("glucose") = 13.97 xx 0.2924 = 4.085 "K"`

∴ Freezing point of glucose solution `("T"_"f") = "T"_"f"^0 - triangle "T"_"f"^'`

= 273.15 K - 4.085 K

= 269.065 K

Answer

Verified

Hint:The depression in freezing point is calculated with the formula \[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\] , \[\Delta {{T}_{f}}\]is the depression in freezing point, \[{{w}_{2}}\] is the mass of the solute in gram, \[{{M}_{2}}\] is the molar mass of the solute and \[{{w}_{1}}\] is mass of the solvent in gram.

Complete step by step answer:

Let us understand about the depression in freezing point:The freezing point of a substance is the temperature at which the solid and the liquid form of the substance are in equilibrium i.e., the solid and liquid form has the same vapor pressure. IIt is seen that the freezing point of the solution is always below the freezing point of the pure solvent. This is called the depression in freezing point.The depression of freezing point depends on the solute in a solution and has found to be related to the molality’ as below:\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }m\]\[{{K}_{f}}\]is the molal depression constant.So, by expanding the above formula, we get:\[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}\]Where, \[\Delta {{T}_{f}}\]is the depression in freezing point,\[{{w}_{2}}\] is the mass of the solute in gram,\[{{M}_{2}}\] is the molar mass of the solute, \[{{w}_{1}}\] is the mass of the solvent in grams.So, according to the question: \[{{w}_{2}}\] is the mass of solute = 68.5 g \[{{M}_{2}}\]= Molar mass of the solute = 342 g / mol\[{{w}_{1}}\]= Mass of the solvent = 1000 gNow, putting all these in the equation, we get \[\Delta {{T}_{f}}={{K}_{f}}\text{ x }\dfrac{{{w}_{2}}}{{{M}_{2}}}\text{ x }\dfrac{1000}{{{w}_{1}}}=1.86\text{ x }\dfrac{68.5}{342}\text{ x }\dfrac{1000}{1000}=0.372\]For calculating the freezing point of the solution we have to subtract the depression in freezing from the freezing point of the pure solvent.\[\begin{align} & \Delta {{T}_{f}}={{T}^{\circ }}-{{T}_{f}} \\ & {{T}_{f}}={{T}^{\circ }}-\Delta {{T}_{f}}=0-0.372=-{{0.372}^{\circ }}C \\ \end{align}\]Since we know that the freezing point of pure water is 0.

Hence the correct option is (b)- \[-{{0.372}^{\circ }}C\].

Note:

\[{{K}_{f}}\] is also called the cryoscopic constant of the solvent. Depression in freezing point is a colligative property because it depends on the number of moles of the solute.

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