Calculate the mass of 95% pure $CaCO_3$ that will be required to neutralize 50 mL of 0.5 M HCl solution according to the following reaction: $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + 2H_2O(l)$ [Calculate up to the second place of decimal point]

At S.T.P. the density of $\text{CCl}_4$ vapour in $\text{g/L}$ will be nearest to:

The molar mass of a compound (X), whose $2.6 \text{ mol}$ weighs $312 \text{ g}$, is:

Calculate the mole fraction of the solute in a $1.00 \text{ m}$ aqueous solution.

A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is: (Given atomic mass of A = 64; B = 40; C = 32 u)

When $50\text{ mL}$ of a $16.9\%\text{ (w/v)}$ solution of $\text{AgNO}_3$ is mixed with $50\text{ mL}$ of $5.8\%\text{ (w/v) NaCl}$ solution, then the mass of precipitate formed is: ($\text{Ag} = 107.8$, $\text{N} = 14$, $\text{O} = 16$, $\text{Na} = 23$, $\text{Cl} = 35.5$)

The density of a $2 \text{ M}$ aqueous solution of $NaOH$ is $1.28 \text{ g/cm}^3$. The molality of the solution is: [molecular mass of $NaOH = 40 \text{ g mol}^{-1}$]

How many moles of lead (II) chloride will be formed from a reaction between $6.5\text{ g}$ of $\text{PbO}$ and $3.2\text{ g}$ of $\text{HCl}$?

If the Avogadro number $N_A$, is changed from $6.022 \times 10^{23} \text{ mol}^{-1}$ to $6.022 \times 10^{20} \text{ mol}^{-1}$ this would change: