Question

Calculate the mass of sodium trioxocarbonate (iv) needed to prepare 2dm³ of 0.9moles

Answer 1

$\begin{aligned} Volume (V)&= 2dm^3\\ Concentration (C)&= 0.9M\\ \textsf{no. of moles of }Na_2CO_3 (n) &= C×V\\ &=0.9× 2 = 1.8 moles \end{aligned}$

$\begin{aligned} \textsf{Molar mass of }Na_2CO_3(M) &= 2(23) + 12 + 3(16)\\ &= 106g/mol \end{aligned}$

$\begin{aligned} \textsf{Mass of }Na_2CO_3&=n× M\\ &= 1.8×106\\ &= 190.8g \end{aligned}$