Q3.6: calculate the number of po43- ions present in 23.7 g of ca3(po4)2.

23.7 g of Ca₃(PO₄)₎₂ contains 9.20 x 10^22 ions of PO₄³⁻  

23.7 g of Ca₃(PO₄)₎₂ contains 0.1528 moles of PO₄³⁻ which is equivalent to 9.20 x 10^22 ions of PO₄³⁻

Therefore;

Ca₃(PO₄)₎₂ ionizes to give out 3 Ca²⁺ ions and 2 PO₄³⁻ ions

Ca₃(PO₄)₎₂ → 3Ca²⁺ + 2PO₄³⁻

Required to determine the number of ions of PO₄³⁻ ions

Number of moles = Mass /Molar mass

Molar mass of Ca₃(PO₄)₎₂ = 310.1767 g/mol

Thus;

Moles = 23.7 g/310.1767 g/mol

          = 0.0764 moles

1 mole of  Ca₃(PO₄)₎₂ ionizes to give 2 moles PO₄³⁻ ions

Therefore;

0.0764 moles will produce;

 = 0.0764 moles x 2

 = 0.0152 moles PO₄³⁻ ions

1 mole of PO₄³⁻ ions contains 6.022 x 10^23 ions

Therefore;

0.01528 moles PO₄³⁻ ions contains;

= 0.01528 x 6.022 x 10^23 ions

= 9.20 x 10^22 ions

Keywords: Atoms, ions, Moles, molar mass

Level: High school

Subject: Chemistry

Topic: Moles

Sub-topic: Avogadro's constant

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol