# A 65.0-mL sample of 0.513 M glucose (C_6H_12O_6) solution is mixed with 140.0 mL of 2.33 M glucose solution. What is the concentration of the final solution?

Jun 6, 2016

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$ $\cong$ $1.7 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

$\text{Moles of solute}$ $=$ $\text{Volume"xx"Concentration}$

$\text{Final concentration}$ $=$ $\text{Total moles of solute (moles)"/"Final volume of solution (Litres)}$ $=$

$\frac{65.0 \times {10}^{-} 3 L \times 0.513 \cdot m o l \cdot {L}^{-} 1 + 140.0 \times {10}^{-} 3 L \times 2.33 \cdot m o l \cdot {L}^{-} 1}{\left(65.0 + 140.0\right) \times {10}^{-} 3 \cdot L}$

$\cong$ $1.7 \cdot m o l \cdot {L}^{-} 1$ with respect to glucose.