Question

the molarity of a 1.5 l solution containing 4 g of mgbr₂ is blank 1 m. **please give your answer to 2 significant figures in the form of 0.xxx. blank 1 add your answer question 14 blank 1 grams of caco₃ is in 4 l of a 3.4 m mgoh solution blank 1 add your answer

Explanation:

Step1: Calculate molar mass of MgBr₂

The molar mass of Mg ($24.31\ g/mol$) and Br ($79.90\ g/mol$). For MgBr₂, molar mass $M = 24.31+2\times79.90=24.31 + 159.8=184.11\ g/mol$.

Step2: Calculate moles of MgBr₂

Use the formula $n=\frac{m}{M}$, where $m = 4\ g$ and $M = 184.11\ g/mol$. So $n=\frac{4}{184.11}\ mol\approx0.0217\ mol$.

Step3: Calculate molarity

Use the formula $Molarity=\frac{n}{V}$, where $n$ is the number of moles and $V = 1.5\ L$. So $Molarity=\frac{0.0217}{1.5}\ mol/L\approx0.014\ M$.

Answer:

0.014