2.5 g of the carbonate of a metal was treated with 100 mL of 1N H2SO4. After the completion of the reaction, the solution was boiled off to expel CO2 and was then titrated against 1N NaOH solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20
Equivalent weight of metal carbonate = 20 + 30 = 50
2.5 gm of metal carbonate = = 0.05 eq.
Number of equivalent of H2SO4 would have reacted = 0.05
Number of equivalent of H2SO4 taken = = 0.1
Number of equivalent weight of H2SO4 reamins unreacted = 0.1 - 0.05 = 0.05 eq.
Number of equivalent of alkali consumed = 0.05 eq.
milli eq = Normality Volume in mL
1.0 V = 0.05 1000
V = = 50 mL
pH of a 0.01 M solution (Ka = 6.6 10-4)
pH of a 0.01 M solution is
pH = 3 - log2.5 = 2.60
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion () of a gas, i.e., = . For an ideal gas is equal to
For n moles of an ideal gas PV = nRT
or V = n
Differentiating with respect to T at constant P, we have
Therefore, for an ideal gas is equal to .
Mercurous chloride exists in the form of
Mercurous chloride or calomel mineral is a rare mineral. It is dense white or yellowisg- white, odorless solid. It is a component of reference electrodes in electrochemistry. It exists in the form of Hg22+.
What is the molarity of H2SO4 solution that has a density of 1.84 g/cc at 35C and contains 98% by weight?
(Since, Volume of solution = )
Therefore, volume of solution = = 54.34 mL
The M - O- M bond angles in MO (where M is halogen) is in the order
Br2O > Cl2O > F2O
F2O > Br2O > Cl2O
F2O > Cl2O > Br2O
Cl2O > F2O > Br2O
Br2O > Cl2O > F2O
In OF2 repulsion between lone pairs is greater than that between bond pair since electrons are away from O and nearer to F. In Cl2O, bonding electrons are nearer to O than to Cl, so the bond angle is greater than 109°28'. In Br2O, the bonding electrons are more closer to oxygen than in Cl2O, so the bond angle is largest (116°).
KF combines with HF to form KHF2. This compound contains the species
K+, F- and H+
K+, F- and HF
K+ and [HF2]-
[KHF]+ and F2
K+ and [HF2]-
The fluoride ion, by a process of hydrogen bonding, forms the anion, HF. The compound is written as K+[HF2]-.
The molecules having the same hybridization, shape and number of lone pairs of electrons are
SeF4 and XeO2F2 are both sp3 hybridized, trigonal bipyramidal and see-saw shaped with 1 lone pair of electrons each.
SF4 has 1 lone pair, XeF2 has 3 lone pairs.
XeOF4 is square pyramidal with 1 lone pair. TeF4 is see-saw shaped with 1 lone pair.
SeCl4 has 1 lone pair and is see-saw shaped. XeF4 has 2 lone pairs and is planar in shape.
The equilibrium constant for mutarotation -D Glucose -D Glucose is 1.8. What percentage of form remains at equilibrium?
Solving = 0.642
1 - = 0.358
Percent of -D glucose remaining at equilibrium = 35.8%
A vessel of one litre capacity containing 1 mole of SO3 is heated till a state of equilibrium is attained.
2SO3 (g) 2SO2 (g) + O2 (g)
At equilibrium, 0.6 moles of SO2 had formed. The value of equilibrium constant is
Equilibrium constant is given by