When 100 mL of 0. 1 M KNO3 and 400 mL of 0.2 M HCI and 500 mL of 0.3 M H2SO4 are mixed, then in the resulting solution

The molarity of K ⊕ = 0.01 M , The molarity of SO 4 2 − = 0.15 M , The molarity of H ⊕ = 0.38 M

When 100 mL of 0. 1 M KNO3 and 400 mL of 0.2 M HCI and 500 mL of 0.3 M H2SO4 are mixed, then in the resulting solution

KNO 3 + HCl + H 2 SO 4 100 × 0 .1 = 10 mmol 400 × 0 .2 = 80 mmol 500 × 0 .3 = 150 mmol Total volume = 100 + 400 + 500 = 1000 mL M K ⊕ = 10 1000 = 0 .01 M M SO 4 − 2 = 150 1000 = 0 .15 M M H ⊕ = 80 + 2 × 150 1000 = 0 .38 M M NO 3 ⊖ = 10 1000 = 0 .01 M M Cl ⊖ = 80 1000 = 0 .08 M

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

When 100 mL of 0. 1 M KNO₃ and 400 mL of 0.2 M HCI and 500 mL of 0.3 M H₂SO₄ are mixed, then in the resulting solution

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The molarity of $K^{\oplus} = 0.01 M$

The molarity of ${SO}_{4}^{2 -} = 0.15 M$

The molarity of $H^{\oplus} = 0.38 M$

The molarity of ${NO}_{3}^{⊖} = 0.08 and {Cl}^{⊖} = 0.01 M$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} {KNO}_{3} + HCl + H_{2} {SO}_{4} \\ (\begin{matrix} 100 \times 0 .1 \\ = 10 mmol \end{matrix}) (\begin{matrix} 400 \times 0 .2 \\ = 80 mmol \end{matrix}) (\begin{matrix} 500 \times 0 .3 \\ = 150 mmol \end{matrix}) \end{array} Total volume = 100 + 400 + 500 = 1000 mL$

$\begin{array}{l} M_{K} \oplus = \frac{10}{1000} = 0 .01 M \\ M_{{SO}_{4}^{- 2}} = \frac{150}{1000} = 0 .15 M \\ M_{H} \oplus = \frac{80 + 2 \times 150}{1000} = 0 .38 M \\ M_{{NO}_{3}}^{⊖} = \frac{10}{1000} = 0 .01 M \\ M_{{Cl}^{⊖}} = \frac{80}{1000} = 0 .08 M \end{array}$