To 50 mL of 0.5 M $H_{2} S O_{4}$ , 75 mL of 0.25 M $H_{2} S O_{4}$ is added. What is the concentration of the final solution if its volume is 125 mL?

1 M

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0.45 M

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0.35 M

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0.15 M

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0.30 M

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Solution

The correct option is A 0.35 M
No. of moles in 0.05 liter of $H_{2} S O_{4}$
= $M \times V (i n l i t e r)$
= $0.5 \times 0.05$
= 0.025

No. of moles in 0.075 liter of $H_{2} S O_{4}$ added
= $0.25 \times 0.075$
= 0.01875

Total no. of moles in 0.125 liter of $H_{2} S O_{4}$
= 0.025 + 0.01875
= 0.04375

∴ Molarity of $H_{2} S O_{4} = \frac{0.04375}{0.0125}$ = $0.35$ M
Answer is C

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To 50 mL of 0.5 M H2SO4, 75 mL of 0.25 M H2SO4 is added. What is the concentration of the final solution if its volume is 125 mL?

The correct option is A 0.35 MNo. of moles in 0.05 liter of H2SO4= M×V(inliter)= 0.5×0.05= 0.025No. of moles in 0.075 liter of H2SO4 added= 0.25×0.075= 0.01875Total no. of moles in 0.125 liter of H2SO4= 0.025 + 0.01875= 0.04375∴ Molarity of H2SO4=0.043750.0125 = 0.35 M Answer is C

To 50 mL of 0.5 M $H_{2} S O_{4}$ , 75 mL of 0.25 M $H_{2} S O_{4}$ is added. What is the concentration of the final solution if its volume is 125 mL?