Question

part a
0.45 mol of na₂so₄ in 7.36 l of solution
express the molarity in moles per liter to two significant figures.
molarity = m
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part b
36.6 g c₂h₆o in 2.04 l of solution
express the molarity in moles per liter to three significant figures.
molarity = m
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Explanation:

Step1: Recall molarity formula

Molarity ($M$) = $\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume of the solution in liters.

Step2: Solve Part A

Given $n = 0.45$ mol and $V=7.36$ L. Using the molarity formula $M=\frac{0.45}{7.36}\approx0.0611$ mol/L. Rounding to two - significant figures, $M = 0.061$ mol/L.

Step3: Calculate molar mass for Part B

The molar mass of $C_2H_6O$: $C$ has an atomic mass of approximately 12 g/mol, $H$ has 1 g/mol and $O$ has 16 g/mol. So, $M_{C_2H_6O}=2\times12 + 6\times1+16=46$ g/mol.

Step4: Calculate moles in Part B

$n=\frac{m}{M}$, where $m = 36.6$ g and $M = 46$ g/mol. So $n=\frac{36.6}{46}=0.79565$ mol.

Step5: Solve for molarity in Part B

Given $V = 2.04$ L, using the molarity formula $M=\frac{n}{V}=\frac{0.79565}{2.04}\approx0.390$ mol/L.

Answer:

Part A: 0.061 M
Part B: 0.390 M