Question

How many milli moles of sucrose should be dissolved in 500 gms of water so that the difference between the boiling point and freezing point of the solution becomes ${103.57}^{o}C$? $(K_f={1.86}^{o}C/m,K_b={0.52}^{o}C/m)$

• A. 500 moles
• B. 90 mmoles
• C. 1330 mmoles
• D. 1330 moles

Answer 1

Answer: (C)

1330 mmoles

The expression for the elevation in the boiling point is as shown below.
$T_{(b,s)}-T_{b,H_{2}O}=K_b\times m.....\left(1\right)$
$T_{(b,s)}$ represents the boiling point of solution. $T_{b,H_{2}O}$ represents the boiling point of water.
The expression for the depression in freezing point is as shown below.
$T_{f,H_{2}O}-T_{f,S}=K_f\times m.....\left(2\right)$
$T_{f,s}$ represents the freezing point of solution. $T_{f,H_{2}O}$ represents the freezing point of water.
Adding equation (1) and equation (2)
$T_{(b,s)}-T_{b,H_{2}O}+(T_{f,H_{2}O}-T_{f,S})=K_b\times m-K_f\times m$
$T_{(b,s)}-T_{f,S}+(T_{f,H_{2}O}-T_{b,H_{2}O})=(K_b-K_f)\times m.....\left(3\right)$
The expression for the molality is, $m=\frac{n}{0.5}.....\left(4\right)$
Substitute values in the equation (3)
$103.57+0-100=(1.86-0.52)\times \frac{n}{0.5}.....\left(3\right)$
Thus, $n=1.33$ mole $=1330$ mmoles