25 ml of 3-0 M HNO 3 are mixed with 75 ml of 4-0 M HNO 3- If the volumes are additive- the molarity of the final mixture would be-A- 3-25 MB- 4-0 MC- 3-75 MD- 3-50 M

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Concentration Terms

25 ml of 3.0 ...

Question

25 ml of 3.0 M HNO₃ are mixed with 75 ml of 4.0 M HNO₃. If the volumes are additive, the molarity of the final mixture would be:

3.25 M

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4.0 M

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3.75 M

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3.50 M

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Solution

The correct option is C 3.75 M
Given: $M_{1} = 3.0 M, V_{1} = 25 m l, M_{2} = 4.0 M, V_{2} = 75 m l$
$M_{3} V_{3} = M_{1} V_{1} + M_{2} V_{2}$
$M_{3} (V_{1} + V_{2}) = M_{1} V_{1} + M_{2} V_{2}$
$M_{3} (25 + 75) = 3 \times 25 + 4 \times 75$
$M_{3} = \frac{375}{100} = 3.75 M$

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25 ml of 3.0 M HNO3 are mixed with 75 ml of 4.0 M HNO3. If the volumes are additive, the molarity of the final mixture would be:

The correct option is C 3.75 MGiven: M1=3.0M,V1=25ml,M2=4.0M,V2=75mlM3V3=M1V1+M2V2M3(V1+V2)=M1V1+M2V2M3(25+75)=3×25+4×75M3=375100=3.75M

25 ml of 3.0 M HNO₃ are mixed with 75 ml of 4.0 M HNO₃. If the volumes are additive, the molarity of the final mixture would be:

The correct option is C 3.75 M
Given: $M_{1} = 3.0 M, V_{1} = 25 m l, M_{2} = 4.0 M, V_{2} = 75 m l$
$M_{3} V_{3} = M_{1} V_{1} + M_{2} V_{2}$
$M_{3} (V_{1} + V_{2}) = M_{1} V_{1} + M_{2} V_{2}$
$M_{3} (25 + 75) = 3 \times 25 + 4 \times 75$
$M_{3} = \frac{375}{100} = 3.75 M$