An open amide of 10g is subjected to Kjeldahl method to estimate the nitrogen. Liberated ammonia is treated with $1 M & 200 m L$ of $H_{2} S O_{4}$ and the left acid is back titrated with 1.5 M, 200 mL $N a O H$ solution. Same quantity of compound when analyzed for carbon & hydrogen gave $63 % o f C & 11 % o f H$ .
[ Note:- After dividing by smallest fraction round off the final number of atoms to nearest whole number] .
Molar mass of the compound $= x (g r m / m o l e)$
Total no. of monomethyl primary amides which give optically active amines on reaction with $B r_{2} / N a O H = y$ ; then $x + y = ?$

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answer is 135.

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Detailed Solution

$% N = \frac{1.4}{10} \times 100 = 14 % C = 63 % H = 11 % O = 12 C \to \frac{63}{12} = 5.25 \to \frac{5.25}{0.75} = 7 H \to \frac{11}{1} = 11 \to \frac{11}{0.75} \approx 15 N \to \frac{14}{14} = 1 \to \frac{1}{0.75} \approx 1 O \to \frac{12}{16} = 0.73 \to \frac{0.75}{0.75} = 1 M . F \to C_{7} H_{15} O N$