An organic compound has an empirical formula $CH_2O$,its vapour density is $45$. The molecular formula of the compound is

$A$ transition metal $M$ forms a volatile chloride which has a vapour density of $94.8$. If it contains $74.75\%$ of chlorine,the formula of the metal chloride will be:

$25 \ g$ of an unknown hydrocarbon upon burning produces $88 \ g$ of $CO_2$ and $9 \ g$ of $H_2O$. This unknown hydrocarbon contains:

$A$ hydrocarbon contains $10.5 \,g$ of carbon and $1 \,g$ of hydrogen. If $2.4 \,g$ of this hydrocarbon occupies a volume of $1 \,L$ at $1 \,atm$ pressure and $127 \,^{\circ}C$ temperature,what is the molecular formula of the hydrocarbon?

Complete combustion of $X$ g of an organic compound gave $0.25$ g of $CO_2$ and $0.12$ g of $H_2O$. If the percentage of carbon is $25\%$ and of hydrogen is $4.89\%$, then $X = \_\_\_\_ \times 10^{-3}$ g. (Nearest integer) (Molar mass of $C$, $H$ and $O$ are $12, 1$ and $16$ g mol$^{-1}$ respectively.)

The value of $n$ in the molecular formula $Be_n Al_2 Si_6 O_{18}$ is