1 mol NaCl in 1000 g $H_{2} O$ will change the boiling point of water to.

{100.51}^{o} C

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{101.02}^{o} C

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{101.53}^{o} C

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{101.86}^{o} C

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{103.6}^{o} C

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Solution

The correct option is B ${101.02}^{o} C$
$△$ $T$ $=$ $i k_{b} m$
for $N a C l$ constant $i$ $=$ $2$ and $K_{b}$ $=$ $.512$
$△$ $T$ $=$ $2 \times 0.512 \times m o l a l i t y$
$m o l a l i t y$ $=$ $w e i g h t / (m o l e c u l a r w e i g h t \times K g s o l v e n t)$
$m o l a l i t y$ $=$ $1 / 1$ $=$ $1$
so $△$ $T$ $=$ $1.024$
So boiling point of this solution $=$ $100 + 1.024$ $=$ $101.02$

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Similar questions

1

N a C l

1000 g

H_{2} O

H_{2} O

K_{b} = {0.52}^{o} C k g / m o l

117 g

N a C l

222 g

Ebullioscopic constant for water (K_{b}) = 0.52 K k g {mol}^{- 1}

Molecular mass of NaCl = 58.5 g m o l^{- 1}

100^{\circ} C

117

_{2}

_{b} =

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