A) 40 g
B) 60 g
C) 100 g
D) 200 g
Correct Answer: A
Solution :
1100 g solution has 60 g urea Water =1040 g \[\Delta {{T}_{f}}=\frac{1000{{K}_{f}}{{w}_{2}}}{{{M}_{2}}{{w}_{1}}}\] \[\frac{\Delta {{T}_{f}}}{{{K}_{f}}}=\frac{1000\times 60}{60\times {{w}_{1}}}=1\] or \[{{w}_{1}}=1000\,g\] The ice formed \[=1040-1000=40\text{ }g\]You need to login to perform this action.
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