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How many milliliters of 0.1087 M Ni(NO₃)² contain 4.079 g of Ni(NO₃)²?

User Aberaud
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Final answer:

To find the volume of 0.1087 M Ni(NO₃)² solution containing 4.079 g of Ni(NO₃)², you need to use the formula Molarity (M) = mol/L. By calculating the number of moles and rearranging the formula, the volume of the solution comes out to be around 0.083 L or 83.0 mL.

Step-by-step explanation:

To calculate the volume of 0.1087 M Ni(NO₃)² solution that contains 4.079 g of Ni(NO₃)², we can use the formula:

Molarity (M) = mol/L

First, we need to calculate the number of moles of Ni(NO₃)² in 4.079 g using its molar mass:

Number of moles = mass/molar mass

Then we can rearrange the formula to solve for volume:

Volume (L) = moles/Molarity

Plugging in the values, we get:

Volume = (4.079 g/ 218.69 g/mol) / 0.1087 mol/L

Simplifying the expression, we find that the volume of the solution is approximately 0.083 L or 83.0 mL.

User Salim B
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