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BigorU [14]
3 years ago
11

). In a titration, a student obtained an average titre value of 3.9 cm3 of 0.3 M HCl. If the volume of Na2CO3 solution used is 1

0 cm3 and the indicator used is phenolphthalein, calculate; (i). the molarity of the Na2CO3 (4 marks) (ii). the mass concentration of the Na2 theCO3. (4 marks)
Chemistry
1 answer:
Oliga [24]3 years ago
5 0

Answer:

I) 0.0585 M

ii)6.2 g dm-3

Explanation:

The reaction equation is given as;

Na2CO3(aq) +2HCl(aq)------> 2NaCl(aq) + CO2(g) +H2O(l)

Concentration of acid CA= 0.3 M

Volume of acid VA= 3.9 cm^3

Concentration of base CB= the unknown

Volume of base VB= 10 cm^3

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB=CBVBNA

CB= CAVANB/VBNA

substituting values;

CB= 0.3 × 3.9 × 1/ 10.0 × 2

CB= 0.0585 M

ii) mass concentration= molar concentration × molar mass

Molar mass of Na2CO3= 106 gmol-1

Mass concentration= 0.0585 × 106 = 6.2 g dm-3

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Olegator [25]

Answer:

1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules

Explanation:

1. Mass / Molar mass = Mol

5g / 28 g/m = 0.178 moles

2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.

4x10²³  x2 = 8x10²³ atoms

3. 1 mol of anything, has 6.02x10²³ particles

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5. Molar mass NaCl = 58.45 g/m

1.3 g /  58.45 g/m = 0.0222 moles

1 mol has 6.02x10²³ atoms

0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²

6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water

Molar mass H₂O = 18 g/m

500 g / 18 g/m = 27.8 moles

6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵

8 0
3 years ago
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