(a) Calculate the molarity of a solution made by dissolving 12.5 grams of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) in enough water to form exactly $750 \mathrm{~mL}\( of solution. (b) How many moles of KBr are present in \)150 \mathrm{~mL}\( of a \)0.112 \mathrm{M}$ solution? (c) How many milliliters of \(6.1 \mathrm{MHCl}\) colution a

1. Calculate the molar mass of Na2CrO4: The molar mass of Na₂CrO₄ can be found by calculating the sum of the molar masses of its individual elements:$$ \mathrm{Na}_{2} \mathrm{CrO}_{4}=2(\mathrm{Na})+\mathrm{Cr}+4(\mathrm{O}) $$Use the atomic masses from the periodic table to find the molar mass of each element:$$ \mathrm{Na} = 22.99\,\mathrm{g/mol} $$ \mathrm{Cr} = 51.996\,\mathrm{g/mol} $$ \mathrm{O} = 16\,\mathrm{g/mol} $$Plug the values into the formula and find the molar mass of Na₂CrO₄:$$ 2(22.99\,\mathrm{g/mol}) + 51.996\,\mathrm{g/mol} + 4(16\,\mathrm{g/mol}) = 161.97\,\mathrm{g/mol} $$2. Convert the mass of Na₂CrO₄ to moles: Now that we have the molar mass, we can convert the given mass (12.5 g) of Na₂CrO₄ to moles:$$ \frac{12.5\,\mathrm{g}}{161.97\,\mathrm{g/mol}}=0.0772\,\mathrm{mol} $$3. Convert volume to liters: Given that the solution has a volume of 750 mL, we need to convert this to liters for the molarity calculation:$$ 750\,\mathrm{mL} \times \frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}=0.750\,\mathrm{L} $$4. Calculate the molarity of the Na₂CrO₄ solution: Plug the values of moles and volume in liters into the molarity formula:$$ M = \frac{n}{V} = \frac{0.0772\,\mathrm{mol}}{0.750\,\mathrm{L}}=0.103\,\mathrm{M} $$Therefore, the molarity of the Na₂CrO₄ solution is 0.103 M.

1. Convert the volume to liters: The given volume of the KBr solution is 150 mL. Convert it to liters:$$ 150\,\mathrm{mL} \times \frac{1\,\mathrm{L}}{1000\,\mathrm{mL}}=0.150\,\mathrm{L} $$2. Calculate the moles of KBr: Given that the molarity of the KBr solution is 0.112 M, use the molarity formula to find the number of moles in the solution:$$ n= M \times V=0.112\,\mathrm{M} \times 0.150\,\mathrm{L} = 0.0168\,\mathrm{mol} $$Therefore, there are 0.0168 moles of KBr in 150 mL of a 0.112 M solution.

It appears that the exercise is incomplete after part (c) because the required volume or moles of HCl are not given. Please provide more information about part (c) to receive a complete solution.