Estimate the number of moles of solute present in each of the following solutions 256 mL of 0.358 M Ca(NO3)2 and 4 liters of 0.516 M HBr??

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Estimate the number of moles of solute present in each of the following solutions 256 mL of 0.358 M Ca(NO3)2 and 4 liters of 0.516 M HBr??

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Answer 1 · 2017-07-31T17:23:57

By definition, $concentration = \frac{moles of solute}{volume of solution}$ $"concentration"="moles of solute"/"volume of solution"$

Explanation:

And thus $moles of solute$ $"moles of solute"$ is given by the product $volume \times concentration$ $"volume"xx"concentration"$ ......

For the $C a {(N O_{3})}_{2} (a q)$ $Ca(NO_3)_2(aq)$ solution, we has........

$256 \cdot m L \times 10^{- 3} \cdot L \cdot m L^{- 1} \times 0.358 \cdot m o l \cdot L^{- 1} = 0.0917 \cdot m o l$ $256*mLxx10^-3*L*mL^-1xx0.358*mol*L^-1=0.0917*mol$ WITH RESPECT TO $calcium nitrate$ $"calcium nitrate"$

And for the $H B r (a q)$ $HBr(aq)$ solution, we has........

$4.00 \cdot L \times 0.516 \cdot m o l \cdot L^{- 1} = 2.064 \cdot m o l$ $4.00*Lxx0.516*mol*L^-1=2.064*mol$ WITH RESPECT TO $hydrobromic acid$ $"hydrobromic acid"$ ......

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Estimate the number of moles of solute present in each of the following solutions 256 mL of 0.358 M Ca(NO3)2 and 4 liters of 0.516 M HBr??

1 Answer

Explanation:

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