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A gaseous compound is composed of 85.7% by mass carbon and 14.3% by mass hydrogen. It's density is 2.28 g/litre at 300 K and 1.0 atm pressure. Determine the molecular formula of the compound:
Calculate the % of free in oleum (a solution of in ) that is labelled 109% .
Percentage above 100 represents the mass of that reacts with dissolved in oleum to give , i.e., 9 g reacts with free to produce
18 g reacts with 80 g
9 g will react with 40 g
or % of free = 40
Suppose two elements X and Y combine to form two compounds and when 0.05 mole of weighs 5 g while 3.0111023 molecules of weighs 85 g. The atomic masses of x and y are respectively:
40 milligram diatomic volatile substance is converted to vapour that displaced 4.92 mL of air at 1 atm and 300 K. Atomic weight of element X is nearly:
For the reaction;
Initially if 2.5 mole of and 3.6 mole of is taken. If 6.3 mole of is obtained then % yield of given reaction is:
0.8 mole of a mixture of CO and CO2 requires exactly 40 gram of NaOH in solution for complete conversion of all the CO2 into Na2CO3. How many moles more of NaOH would it require for conversion into Na2CO3, if mixture (0.8 mole) is completely oxidised to CO2?
The impure 6 g of is dissolved in water and then treated with excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be 14 g. The % purity of solution would be:
25.4 of and 14.2 g of are made to react completely to yield a mixture of ICI and ICI3. Calculate mole of ICI and ICI3 formed.
The vapour density of a mixture containing and is 38.3. Calculate the mole of in 100 g mixture.
Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides?
Let valencies of Cu in two oxides be x and y, then
I oxides is
II oxides is
In I oxide:
Equivalent of Cu = Equivalent of oxygen
where w, A, x and a are mass of Cu, at. mass of Cu, valency of Cu and mass of oxygen
In II oxide:
By Eqs. (i) and (ii)
Valency of Cu in I and II oxides are in the ratio 2:1
5 mL of a gaseous hydrocarbon was exposed to 30 mL of . The resultant gas, on cooling is found to measure 25 mL of which 10 mL is absorbed by NaOH and the remainder by pyrogallol. Determine molecular formula of hydrocarbon. All measurements are made at constant room temperature.
Suppose formula of hydrocarbon is
Volume taken initially
Volume after reaction
Given that, left is absorbed by pyrogallol
Also volume absorbed by NaOH = 10 mL = Volume of formed
Total volume of gases left after reaction = 25 mL
or Volume of left + Volume of formed = 25 mL
Volume of left = 25-10 =15
30-5 [a+(b/4)]=15 b=4
Thus, hydrocarbon is .
A sample of and weighed 2.21 g is ignited to constant mass of 1.152 g. What is the composition of mixture? Also calculate the volume of evolved at and 76 cm of pressure.
The vapour density of a volume chloride of a metal is 95 and the specific heat of the metal is 0.13 cal/g. The equivalent mass of the metal will be:
Which mixture is lighter than humid air?
Humid air is lighter than dry air as its average molar mass is less. Presence of in air gives lower molar mass than humid air. has lower molar mass than .
is absorbed by:
reacts with turpentine oil or oil of cinnamon.