# 516 mL of a 3.82 M sodium sulfate (Na_2SO_4) solution is diluted with 0.875 L of water. What is the new concentration in molarity?

Apr 26, 2016

Approx. $1.4 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

$\text{Molarity "= " Moles of solute"/"Volume of solution (L)}$

$=$ $\frac{516 \times {10}^{-} 3 \cdot \cancel{L} \times 3.82 \cdot m o l \cdot \cancel{{L}^{-} 1}}{0.516 \cdot L + 0.875 \cdot L}$

$=$ $\frac{1.97 \cdot m o l}{1.391 \cdot L}$

$\cong$ $1.4 \cdot m o l \cdot {L}^{-} 1$