$1.00 g$ of a non-electrolyte solute dissolved in $50 g$ of benzene lowered the freezing point of benzene by $0.40 K$ . The freezing point depression constant of benzene is $5.12 K k g m o l^{- 1}$ . Find the molar mass of the solute.

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Solution

Given:
Mass of non-electrolyte solute $(m_{2}) = 1.00 g$ ,
Mass of benzene solvent $(m_{1}) = 50 g$
Decrease in freezing point $Δ T_{f}^{0} = 0.4 K$
And freezing point depression constant of benzene $(K_{f}) = 5.12 K k g m o l^{- 1}$

Molar mass of solute

We know that $Δ T_{f} = K_{f} \times m o l a l i t y (m)$
We also know that molality = $\frac{mass of solute}{molar mass of solute \times mass of solvent(kg)}$

So,
$Δ T_{f} = K_{f} \times \frac{mass of solute(m_2)}{molar mass of solute(M_2) \times mass of solvent(m_1)}$

So, $M_{2} = K_{f} \times \frac{mass of solute(m_2)}{Δ T_{f} \times mass of solvent(m_1)}$

$M_{2} = \frac{5.12 K k g m o l^{- 1} \times 1.00 g}{0.4 K \times 50 \times 10^{- 3} k g} = 256 g m o l^{- 1}$

Thus, molar mass of the solute $= 256 g m o l^{- 1}$

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1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol−1. Find the molar mass of the solute.

$1.00 g$ of a non-electrolyte solute dissolved in $50 g$ of benzene lowered the freezing point of benzene by $0.40 K$ . The freezing point depression constant of benzene is $5.12 K k g m o l^{- 1}$ . Find the molar mass of the solute.