0.44 g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at S.T.P., `112 cm^(3)` of methane. With PCC the same alcohol forms a carbonyl compound that answers silver mirror test. The monohydric alcohol is

Options

A. 0.01 M NaCl
B. `0.01 " M "Na_(2)SO_(4)`
C. 0.1 M Sucrose
D. 0.1 M NaCl

Answer

A

Explanation

Depression in frezzing point , `Delta T_(f) = I K_(f) m DeltaT_(f)` will follow the order :
`{:( 0.1 M NaCl ,gt ,0.1 M Sucrose , gt, 0.01 M Na_(2)SO_(4) gt),(i=2,,i=1,,i=3),(,,,,0.01" M "NaCl),(,,,,i = 2 ):} `
Depression in freezing point is minimum in case of 0.01 M NaCl solution hence , it will have maximum freezing point .