A solution of 3.0 g of a non-volatile non-electrolyte in 90 g of water shows a relative lowering of vapour pressure of 3.322x10^-3. Using Raoult's law exactly (molar mass of water 18), the molar mass of the solute in g mol^-1 is

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A solution of $3.0 g$ of a non-volatile non-electrolyte in $90 g$ of water shows a relative lowering of vapour pressure of $3.322 \times 1 0^{- 3}$ . Using Raoult's law exactly (molar mass of water $18$ ), the molar mass of the solute in $g mo l^{- 1}$ is

x_{2} = n_{2} / (n_{1} + n_{2}) = 3.322 \times 1 0^{- 3}

with

n_{1} = 90/18 = 5.00

. Then

n_{2} = x_{2} n_{1} / (1 - x_{2}) = 0.01667 mol

M_{2} = 3.0/0.01667 = 180.0 g mo l^{- 1}