A solution was prepared by dissolving 66.6 g of calcium chloride in enough water to make 175 mL of solution. This solution was then further diluted to a total of 550 mL with water. What was the concentration of the final solution?

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A solution was prepared by dissolving 66.6 g of calcium chloride in enough water to make 175 mL of solution. This solution was then further diluted to a total of 550 mL with water. What was the concentration of the final solution?

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Answer 1 · 2016-04-16T12:59:47

$Concentration = \frac{Moles of solute}{Volume of solution}$ $"Concentration "="Moles of solute"/"Volume of solution"$ , $approx. 1 \cdot m o l \cdot L^{- 1}$ $"approx. "1*mol*L^-1$

Explanation:

$C$ $C$ $=$ $=$ $\frac{n}{V}$ $n/V$ , where $n = \frac{66.6 \cdot g}{110.98 \cdot g \cdot m o l^{- 1}}$ $n=(66.6*g)/(110.98*g*mol^-1)$ $=$ $=$ $0.600$ $0.600$ $m o l$ $mol$ .

The final volume of the solution is $550 \times 10^{- 3} \cdot L$ $550xx10^-3*L$

So $Concentration with respect to C a C l_{2} = \frac{0.600 \cdot m o l}{550 \times 10^{- 3} \cdot L}$ $"Concentration with respect to "CaCl_2=(0.600*mol)/(550xx10^-3*L)$ $=$ $=$ $? ?$ $??$

What is the concentration with respect to $C a^{2 +} (a q)$ $Ca^(2+)(aq)$ , to $C l^{-} (a q)$ $Cl^(-)(aq)$ ?

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A solution was prepared by dissolving 66.6 g of calcium chloride in enough water to make 175 mL of solution. This solution was then further diluted to a total of 550 mL with water. What was the concentration of the final solution?

1 Answer

Explanation:

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