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The molal freezing point constant of wat...

The molal freezing point constant of water is `1.86 K m^(-1)`. If `342 g` of cane sugar `(C_(12)H_(22)O_(11))` is dissolved in `1000 g` of water, the solution will freeze at

A

`-1.86 ^@C`

B

`1.86^@C`

C

`-3.92^@C`

D

`2.42^@C`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • The molal freezing points constant of water is 1.86 K kg "mol"^(-1) . Therefore, the freezing point of 0.1 M NaCl in water is expected to be 1) -1.86^(@)C 2) -0.372^(@)C 3) -0.186^(@)C 4) +0.372^(@)C

    A
    `-1.86^(@)C`
    B
    `-0.372^(@)C`
    C
    `-0.186^(@)C`
    D
    `+0.372^(@)C`

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