Question

how many grams of potassium carbonate are needed to make 400.0 ml of 2.5 m solution?
k₂co₃: 138.21 g/mol
86 g k₂co₃ 140,000 g k₂co₃
21 g k₂co₃ 140 g k₂co₃

Explanation:

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) per liter of solution ($V$ in liters), so $M=\frac{n}{V}$. We need to find moles first. The volume is $400.0\space mL = 0.4000\space L$, and $M = 2.5\space M$. Rearranging the formula, $n = M\times V$.
$n = 2.5\space mol/L\times0.4000\space L = 1.0\space mol$

Step2: Calculate Mass from Moles

Mass ($m$) is moles ($n$) times molar mass ($MM$). The molar mass of $\ce{K2CO3}$ is $138.21\space g/mol$. So $m = n\times MM$.
$m = 1.0\space mol\times138.21\space g/mol = 138.21\space g\approx140\space g$ (considering significant figures and answer choices)

Answer:

D. 140 g $\ce{K2CO3}$ (assuming the last option is D, with text "140 g $\ce{K2CO3}$")