Question

How would you prepare from the solid and pure water (a) \(0.400 \mathrm{~L}\) of \(0.155 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) ? (b) \(1.75 \mathrm{~L}\) of \(0.333 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} ?\)

Short answer

Answer: Approximately 9.65 grams of solid Sr(OH)2.

Step-by-step solution

Calculate moles of Sr(OH)2

First, we need to determine the moles of Sr(OH)2 required to make 0.400 L of 0.155 M Sr(OH)2 solution. moles of Sr(OH)2 = (0.155 M) × (0.400 L) moles of Sr(OH)2 = 0.0620 mol

Calculate mass of Sr(OH)2

Now, we'll find the mass of Sr(OH)2 required by multiplying the moles by its molar mass. Molar mass of Sr(OH)2 = 121.63 g/mol (Sr) + 2 × (15.999 g/mol (O) + 1.008 g/mol (H)) Molar mass of Sr(OH)2 = 121.63 + 2 × (15.999 + 1.008) Molar mass of Sr(OH)2 = 121.63 + 2 × 17.007 Molar mass of Sr(OH)2 = 121.63 + 34.014 Molar mass of Sr(OH)2 = 155.644 g/mol mass of Sr(OH)2 = (0.0620 mol) × (155.644 g/mol) mass of Sr(OH)2 ≈ 9.65 g

Calculate moles of (NH4)2CO3

Next, we need to determine the moles of (NH4)2CO3 required to make 1.75 L of 0.333 M (NH4)2CO3 solution. moles of (NH4)2CO3 = (0.333 M) × (1.75 L) moles of (NH4)2CO3 ≈ 0.58275 mol

Calculate mass of (NH4)2CO3

Now, we'll find the mass of (NH4)2CO3 required by multiplying the moles by its molar mass. Molar mass of (NH4)2CO3 = 2 × (1.008 g/mol (N) + 4 × 1.008 g/mol (H)) + 12.011 g/mol (C) + 3 × 15.999 g/mol (O) Molar mass of (NH4)2CO3 = 2 × (14.007) + 12.011 + 3 × 15.999 Molar mass of (NH4)2CO3 = 28.014 + 12.011 + 3 × 15.999 Molar mass of (NH4)2CO3 = 28.014 + 12.011 + 47.997 Molar mass of (NH4)2CO3 = 96.026 g/mol mass of (NH4)2CO3 = (0.58275 mol) × (96.026 g/mol) mass of (NH4)2CO3 ≈ 56.00 g To prepare the solutions: (a) Weigh approximately 9.65 g of solid Sr(OH)2 and dissolve it in pure water to obtain 0.400 L of 0.155 M Sr(OH)2 solution. (b) Weigh approximately 56.00 g of solid (NH4)2CO3 and dissolve it in pure water to obtain 1.75 L of 0.333 M (NH4)2CO3 solution.