c
Let \(M\) is the molar mass of the compound \(( g / mol )\) mass of compound \(=0.01\,M\,gm\)

\(\text { mass of carbon }=0.01 M \times \frac{60}{100}\)

\(\text { moles of carbon }=\frac{0.01 M }{12} \times \frac{60}{100}\)

moles of \(CO _2\) from combustion \(=\frac{4.4}{44}=\) moles of carbon

\(\frac{0.01\,M }{12} \times \frac{60}{100}=\frac{4.4}{44}\)

\(M =\frac{4.4}{44} \times \frac{100}{60} \times \frac{12}{0.01}=200\, gm / mol\)