The molal freezing point constant for water is 1.86 K/m. Therefore, the depression in freezing point of 0.1 M $N a C l$ solution in water is expected to be:

- {1.86}^{0} C

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- {0.186}^{0} C

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- {0.372}^{0} C

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+ {0.372}^{0} C

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Solution

The correct option is D $+ {0.372}^{0} C$
$K_{f} = 1.8 K / m$
$Δ T_{f} = i K_{f} \times m$
$= 2 \times 1.86 \times 0.1$
$= {0.372}^{o} C$

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1.86 C / M

0.1 M N a C l

K_f for water is 1.86Kkg/molal. Therefore the freezing point of 0.1M NaCl solution is:

The freezing point of one molal NaCl solution assuming NaCl to be 100% dissociated in water is (molal depression constant = 1.86)

0.5 m

K C l

{1.86}^{o} C m^{- 1}

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