Molar Volume
One mole of any gas at STP will have a volume of 22.4 L. This is called molar volume. The molar volume [22.4 L at STP] plays a vital role in stoichiometric calculations because it is the link between volume and mass in reactions ..
One mole of any gas at STP will have a volume of 22.4 L. This is called molar volume. The molar volume [22.4 L at STP] plays a vital role in stoichiometric calculations because it is the link between volume and mass in reactions ..Molarity
of the solution is defined as the number of moles of solute dissolved per litre (dm 3 ) of the solution. It is denoted by the symbol M. Measurements in Molarity can change with the change in temperature because solutions expand or contract accordingly. In terms of weight, mol..
of the solution is defined as the number of moles of solute dissolved per litre (dm 3 ) of the solution. It is denoted by the symbol M. Measurements in Molarity can change with the change in temperature because solutions expand or contract accordingly. In terms of weight, mol..Molarity equation
To calculate the volume of a definite solution required to prepare solution of other molarity, the following equation is used: M 1 V 1 = M 2 V 2 , where M 1 = initial molarity, M 2 = molarity of the new solution, V 1 = initial volume and..
Molar Volume
Molar Volume - Molar volume is the volume occupied by one mole of any gas at a definite pressure and temperature. It is denoted by V m . Molar volume of the substance depends on temperature and pressure. The unit of molar..
Standard Molar Volume
Standard molar volume of a gas is the volume occupied by 1 mole of any gas at 273 K and 1 atm pressure (STP). It is equal to 22.4 litres of 22,400 ml. It is the same for all gases. Remember S.T.P. = Standard Temperature and Pressure Standard Temperature = 0 o C or..
Molar volume Animation
Mole Concept and Stoichiometry..
Mole Concept and Stoichiometry..Calculation
Mass of U-tube with contents before experiment = a grams Mass of U-tube with contents after experiment = b grams Hence, mass of water vapour absorbed = (b - a) g Volume of air = x litres..
Mass of U-tube with contents before experiment = a grams Mass of U-tube with contents after experiment = b grams Hence, mass of water vapour absorbed = (b - a) g Volume of air = x litres..Calculations
Let the mass of platinic chloride salt be = W g The mass of platinum residue left = x g From the molar stoichiometry, Number of mol of B 2 H 2 PtCl 6 = Number of mol of Pt Or If E is the equivalent mass of the organic base, Then molecular mass of the salt (B 2 H 2 PtCl 6 ) = 2E + 2 + ..
Let the mass of platinic chloride salt be = W g The mass of platinum residue left = x g From the molar stoichiometry, Number of mol of B 2 H 2 PtCl 6 = Number of mol of Pt Or If E is the equivalent mass of the organic base, Then molecular mass of the salt (B 2 H 2 PtCl 6 ) = 2E + 2 + ..Calculations
Let, The mass of the acid dissolved be = W g Volume of alkali used for complete neutralization of acid solution = V 1 mL Normality of the alkali solution = N 1 Then, we can write, V 1 mL of alkali solution having normality N 1 = Wg acid From the definition, 1000 mL of 1N alkali so..
Let, The mass of the acid dissolved be = W g Volume of alkali used for complete neutralization of acid solution = V 1 mL Normality of the alkali solution = N 1 Then, we can write, V 1 mL of alkali solution having normality N 1 = Wg acid From the definition, 1000 mL of 1N alkali so..See what our Users say :
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