Question

# A $5%$ solution of cane sugar with a molecular weight of $342$ is isotonic with a $1%$ solution of a substance $X$. What is the molecular weight of $X$?

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Solution

## Step 1: Given data${W}_{\mathit{1}}=5g$${\mathrm{V}}_{1}=100\mathrm{ml}=0.1\mathrm{l}$${\mathrm{M}}_{1}=342$$X$:${W}_{\mathit{2}}=1g$${\mathrm{V}}_{2}=100\mathrm{ml}=0.1\mathrm{l}$${M}_{\mathit{2}}=?$Step 2: We know that, for an isotonic solution, ${C}_{\mathit{1}}={C}_{\mathit{2}}$$\frac{{W}_{\mathit{1}}}{{M}_{\mathit{1}}{V}_{\mathit{1}}}=\frac{{W}_{\mathit{2}}}{{M}_{\mathit{2}}{V}_{\mathit{2}}}$ Step 3: Substituting the values.$\frac{5}{\left(341×0.1\right)}=\frac{1}{\left(M2x0.1\right)}\phantom{\rule{0ex}{0ex}}{M}_{\mathit{2}}=\frac{342}{5}\phantom{\rule{0ex}{0ex}}{M}_{\mathit{2}}=68.4\mathrm{g}$

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