The answer is 17 and a half
Changing the volume increases the area that the molecules collide with so the force is spread over a larger area.
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to 
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is 
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Element Atomic Number Valency
Valency of Hydrogen 1 1
Valency of Helium 2 0
Valency of Lithium 3 1
Valency of Beryllium 4 2
Valency of Boron 5 3
Valency of Carbon 6 4
Valency of Nitrogen 7 3
Valency of Oxygen 8 2
Valency of Fluorine 9 1
Valency of Neon 10 0
Valency of Sodium (Na) 11 1
Valency of Magnesium (Mg) 12 2
Valency of Aluminium 13 3
Valency of Silicon 14 4
Valency of Phosphorus 15 3
Valency of Sulphur 16 2
Valency of Chlorine 17 1
Valency of Argon 18 0
Valency of Potassium (K) 19 1
Valency of Calcium 20 2
Valency of Scandium 21 3
Valency of Titanium 22 4
Valency of Vanadium 23 5,4
Valency of Chromium 24 2
Valency of Manganese 25 7, 4, 2
Valency of Iron (Fe) 26 2, 3
Valency of Cobalt 27 3, 2
Valency of Nickel 28 2
Valency of Copper (Cu) 29 2, 1
Valency of Zinc 30 2
Answer is: mass of <span>potassium bromide is 4.71 grams.
V(KBr) = 25.4 mL </span>÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L 0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.
