Question

In front of you is a solid cube measuring 39.68 mm long, 31.68 mm wide, and 12.82 mm high. The cube has a density of 7.586 g/cm3 and is 29.53% cesium phosphate by mass. Next to the cube is a beaker containing 28.57 in3 of 0.1571 M solution of mercury (I) nitrate. How many formula units of solid will be formed when the cube is placed in the beaker? (Assume that the part of the cube that is not cesium phosphate is soluble and does not react.

Answer 1

length of cube = 39.68 mm = 3.968 cm

width of cube = 31.68 mm = 3.168 cm

height of cube = 12.82 mm = 1.282 cm

volume of cube = (length of cube) * (width of cube) * (height of cube)

volume of cube = (3.968 cm) * (3.168 cm) * (1.282 cm)

volume of cube = 16.1 cm³

mass of cube = (volume of cube) * (density of cube)

mass of cube = (16.1 cm³) * (7.586 g/cm³)

mass of cube = 122.3 g

mass cesium phosphate = (% cesium phosphate) * (mass of cube)

mass cesium phosphate = (29.53 %) * (122.3 g)

mass cesium phosphate = (29.53 / 100) * (122.3 g)

mass cesium phosphate = 36.10 g

moles cesium phosphate = (mass cesium phosphate) / (molar mass cesium phosphate)

moles cesium phosphate = (36.10 g) / (493.69 g/mol)

moles cesium phosphate = 0.0731255 mol

volume of mercury (I) nitrate = 28.57 in³

volume of mercury (I) nitrate = 28.57 in³ * (1 L / 61.024 in³)

volume of mercury (I) nitrate = 0.4682 L

moles mercury (I) nitrate = (molarity mercury (I) nitrate) * (volume of mercury (I) nitrate)

moles mercury (I) nitrate = (0.1571 M) * (0.4682 L)

moles mercury (I) nitrate = 0.07355 mol

The balanced reaction is : 3 HgNO₃ (aq) + Cs₃PO₄ (aq) Hg₃PO₄ (s) + 3 CsNO₃ (aq)

moles Hg₃PO₄ formed = (1/3) * (moles mercury (I) nitrate)

moles Hg₃PO₄ formed = (0.333) * (0.07355 mol)

moles Hg₃PO₄ formed = 0.02452 mol

formula units of Hg₃PO₄ formed = (moles Hg₃PO₄ formed) * (Avogadro's number)

formula units of Hg₃PO₄ formed = (0.02452 mol) * (6.022 x 10²³ formula units/mol)

formula units of Hg₃PO₄ formed = 1.476 x 10²² formula units