The following solutions are mixed (volumes additive): 150 mL of 0.20 M NaCl, 250 mL of 0.10 M CaCl2 and 100 mL of 0.30 M AlCl3. Assuming complete dissociation, calculate the molarity of Cl- in the final mixture (two decimal places).

JEE Advanced levelNumerical answerDifficulty 3 of 3so148

The following solutions are mixed (volumes additive): $150 mL$ of $0.20 M$ $NaCl$ , $250 mL$ of $0.10 M$ $CaCl X_{2}$ and $100 mL$ of $0.30 M$ $AlCl X_{3}$ . Assuming complete dissociation, calculate the molarity of $Cl X^{-}$ in the final mixture (two decimal places).

Numerical answer type, answer in M, tolerance 0.01.

Moles of

Cl X^{-}

0.150 \times 0.20 = 0.030

;

0.250 \times 0.10 \times 2 = 0.050

;

0.100 \times 0.30 \times 3 = 0.090

. Total

= 0.170

mol in

0.500

L. So

[Cl X^{-}] = 0.170/0.500 = 0.34