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Answer the following questions assuming you have a 1.3M (molarity) solution ofHNO3 (5 points each, 15 points total)A. How many moles of HNO3 would you have in 0.6 liters of the above solution?B. How many grams of HNO3 would be present in 0.6 liters of the above solution?C. If you began with 0.6 liters of solution, and diluted the solution to a concentration of 0.45M, whatwould the final volume of the solution be?

Respuesta :

Answer:

A) 0.78 moles

B) 49.15grams

C) 1.73L

Explanations:

The formula for calculating the molarity of a solution is expressed as;

[tex]\begin{gathered} molarity=\frac{moles}{volume} \\ moles=molarity\times volume \end{gathered}[/tex]

Given the following parameters

molarity of HNO3 = 1.3M

volume of HNO3 = 0.6L

A) moles of HNO3 = 1.3 * 0.6

moles of HNO3 = 0.78moles

B) Mass of HNO3 = moles * molar mass

Mass of HNO3 = 0.78 * 63.01

Mass of HNO3 = 49.15grams

There are 49.15grams of HNO3 present in 0.6L of the solution

C) According to dilution formula

[tex]\begin{gathered} C_1V_1=C_2V_2 \\ V_2=\frac{C_1V_1}{C_2} \\ V_2=\frac{1.3\times0.6}{0.45} \\ V_2=\frac{0.78}{0.45} \\ V_2=1.73L \end{gathered}[/tex]

Therefore the final volume of the solution will be 1.73L

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