Molarity of HCl solution
20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is: (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.
20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is: (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9
Choose the right answer from above.
Expert
The right answer is (i)
M1V1 = M2V20.01 x 19.85 = M2 x 20M2= 0.00925 : M : 0.0099
The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea (b) BaCl2 > AlCl3 > Urea (c) Urea > BaCl2<
Give me answer of this question. The formula weight of H2SO4 is 98. The weight of the acid in 400mi of solution is: (a)2.45g (b) 3.92g (c) 4.90g (d) 9.8g
A) It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry out the following calculation to test this hypothesis. Suppose that the width of the skate in contact with the ice has been reduced by sh
Write a short note on the IUPAC name of the benzene?
Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal
what is the equation for calculating molar mass of non volatile solute
When 5.85 g of NaCl (having molecular weight 58.5) is dissolved in water and the solution is prepared to 0.5 litres, the molarity of the solution is: (i) 0.2 (ii) 0.4 (iii) 1.0 (iv) 0.1
Help me to go through this problem. 6.02x 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is: (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M (Avogadro constant, N4= 6.02x 1023mol -1)<
lculwhat is the equation for caating molar mass of non volatile solute
The quantum mechanical methods, illustrated previously by the Schrödinger equation, are extended by the use of operators. Or, w
18,76,764
1954680 Asked
3,689
Active Tutors
1414994
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!