Calculate the amount of heat released when:
(i) 100 mL of 0.2 M HCl solution is mixed with 50 mL of 0.2 M KOH.
(ii) 200 mL of 0.1 M `H_(2)SO_(4)` is mixed with 200 mL of 0.2 M KOH solution.
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calcualate the amount of heat evolved when `500 cm^(3)` of 0.1M HCl is mixed with `200 cm^(3)` of 0.2 M NaOH.
Correct Answer - c `HCl+NaOHto NaCl+H_(2)O` at t=0, number of moles = `(500xx0.1)/1000 = (200xx0.2)/1000 ` =0.05 =0.04 during neutralisation of 1 mole of NaOH by 1 mole of HCl, heat...
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Assertion(A) : Heat of neutralisation of `HCl` and `NaOH` is same as that of `H_(2)SO_(4)` with `NaOH`. Reason(R ) : `HCl`, `H_(2)SO_(4)` and `NaOH` a
Correct Answer - A Heat of neutralisation of strong acid with strong base is contant `(-137 Kcal//mol^(-1))`
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Upon titrating `50mL` of `1.96%(w//v) H_(2)SO_(4)` with a `KOH` solution (containing `11.2g KOH` per litre of solution on adding `50mL KOH` solution.
`[H_(2)SO_(4)]=(19.6)/(98)=0.2 M` and `KOH=(11.2)/(56)=0.2M` so for equivalent point, `(0.2xx50xx2)/(0.2)=100mL` of `KOH`solution should be added. `[H^(+)]=(0.2xx2-0.2)/(2)=0.1M` so `pH=1`.
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When 100 mL each of HCl and NaOH solutions are mixed, 5.71 kJ of heat was evolved. What is the molarity of two solution? The heat of neutralisation of
Correct Answer - 1M
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The amount of heat released when `20 ml`,`0.5 M` `NaOH`is mixed with `100 ml`,`0.1 M` `H_(2)SO_(4)` is `xKJ`. The heat of neutralization will be `:-`
Correct Answer - A Number of moles of `NaOH=(MV)/(1000)=(0.5xx20)/(1000)=0.01` Number of moles of `HCl=(MV)/(1000)=(0.1xx100)/(1000)=0.01` Heat of neutralization `=(-x)/(0.01)=-100x`
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`150 mL` of `0.5N HCl` solution at `25^(@)C` was mixed with `150 mL` of `0.5 N NaOH` solution at same temperature. Calculate the heat of neutralizatio
Correct Answer - `-16 kcal`.
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Sulphur combines with oxygen to form two oxides `SO_(2)` and `SO_(3)` if 10 g of S is mixed with 12 g of `O_(2)`. The mole of `SO_(2)` and `SO_(3)` wi
Correct Answer - 6 Li, P, C, F, Cl, S
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If 1 litre of 9.8% w/w `H_(2)SO_(4)` (d = 1.5 g/ml) solution is mixed with 4 litre of 1M KOH solution then pH of resultant solution will be given (log
Correct Answer - A Stability of carbocation by resonance of benzene
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Prove that `^100 C_2^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[^(200)C_(98)-^(100)C_(49)]dot`
To find `S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)` Consider, `.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)` `= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)` = Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)` `= .^(200)C_(98) " "(1)` Also,...
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The value of `((100),(0))((200),(150))+((100),(1))((200),(151))+......+((100),(50))((200),(200))` equals (where `((n),(r ))="^(n)C_(r)`)
Correct Answer - A `(a)` Writing `((200),(r ))=((200),(200-r))`, we have `((100),(0))((200),(50))+((100),(1))((200),(49))+((100),(2))((200),(48))+......+((100),(50))((200),(0))` `=` Coefficient of `x^(50)` in the expansion of `(1+x)^(100)(x+1)^(200)` `=` Coefficient of `x^(50)` in the binomial expansion of `(1+x)^(300)` `=((300),(50))`
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