Question

determine the mass(g) needed to make 850.0 ml of a 0.0650 m solution of tungsten(iv) permanganate. - use correct significant digits and no unit -

Explanation:

Step1: Convert volume to liters

$V = 850.0\ \text{mL} = \frac{850.0}{1000} = 0.8500\ \text{L}$

Step2: Calculate moles of solute

Moles $n = M \times V = 0.0650\ \text{mol/L} \times 0.8500\ \text{L} = 0.05525\ \text{mol}$

Step3: Find molar mass of compound

Tungsten(IV) permanganate is $\text{W(MnO}_4\text{)}_4$.
Molar mass:
$M_W = 183.84\ \text{g/mol}$, $M_{Mn} = 54.94\ \text{g/mol}$, $M_O = 16.00\ \text{g/mol}$
$M_{\text{W(MnO}_4\text{)}_4} = 183.84 + 4\times(54.94 + 4\times16.00) = 183.84 + 4\times(54.94+64.00) = 183.84 + 4\times118.94 = 183.84 + 475.76 = 659.6\ \text{g/mol}$

Step4: Calculate mass of solute

$m = n \times M = 0.05525\ \text{mol} \times 659.6\ \text{g/mol} \approx 36.4$

Answer:

36.4