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 HCl is 57.1 kJ
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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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What volume of aqueous solution of NaOH that is `80%` by mass NaOH, contanis 0.4 mol of NaOH. Density of solution is `0.8 gm//ml`.
Correct Answer - B `0.4 " mole " NaOH = 0.4 xx 40 = 16 gm` `80%` by mass NaOH `80 gm NaOH` in 100 gm solution 16 gm NaOH in...
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4.0 g of NaOH are present in one decilitre of solution . Calcute Mole fraction of NaOH Molality of solution Molarity of solution.
Calculation of molarity of the solution Mass of NaOH =4g, Molar mass of NaoH=40 g `mol^(-1)` Volume of solution=1 decilitre=0.1L Molarity (M)`=("Mass of NaOH"//"Molar mass of NaOH")/("Volme of solution in...
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2.5 litre of 1M NaOH solution is mixed with a 3.0 litres of 0.5 M NaOH solution. The molarity of the resulting solution is :
Correct Answer - c `M_(1)V_(1)+M_(2)V_(2)=M_(3)(V_(1)+V_(2))` `M_(3)=(M_(1)V_(1)+M_(2)V_(2))/((V_(1)+V_(2)))` `M_(3)=(1xx2.5+0.5xx3)/(2.5+3)` `(4.0)/(5.5)=0.73M`
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To a 100 mL of 0.1 M weak acid HA solution 22.5 mL of 0.2 M solution of NaOH are added.Now, what volume of 0.1 M NaOH solution be added into above sol
Correct Answer - A `HA+NaOHtoNaA+H_2O` `t=0 " " 10 " " 4.5 " " 0 " " 0` t=50% `" " 5 " " 0 " " 5` 5 mL NaOH...
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300 gm, `30%`(w/w) NaOH solution is mixed with 500 gm `40%`(w/w) NaOH solution. What is `%` (w/v) NaOH if density of final solution is 2 gm/mL?
Correct Answer - A % (w/v) volume of NaOH`=((300xx0.3+500xx0.4)/(800))/2xx100=72.5`
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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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`100 cm^(3)` of `0.5N HCI` solution at `299.95 K` was mixed with `100 cm^(3)` of ` 0.5 N NaOH` solution at `299.75 K` in a thermos flask. The final te
The initial average temperature of the acid and the `=(299.95+299.75)/(2)=299.85K` rise in temperature `=(302.65-299.65)=2.80K` Heat evolved during neutralisation `=(100+1000+44)xx4.184xx2.8=2858.5J` `therefore` Enthalpy of neutralisation `=-(2858.5)/(100)xx1000xx(1)/(0.50)` `=-57.17kJ`
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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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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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