1. What amount of solute is present in each of the following solutions? a. 750 m

Question

1. What amount of solute is present in each of the following solutions? a. 750 ml of 2.00 M HNOs (1.50 moles, 94.5g) b. 450 ml of 1.25 M Ci2H22011 (0.562 moles, 192g) 2 whatis the (1) equivalent weight, (2) # of equivalents, and (3) normality of each of the following? a. 0.800g NaOH in 500 ml (40.0g, 0.0200 equiv, 0.0400N) b. 18.5g Ba(0H)2 in 600 ml (85.5g, 0.216 equiv, 0.360N) 3. Calculate the normality of NaOH solutions which require: a. 42.0 ml to neutralize 22.0 ml of 0.200N HCI (0.105N) b. 21.8 ml to neutralize 26.0 ml of 0.750 N H2SO4 (0.894N) 4. What is the concentration of each of the following? a. 15.0 ml of 0.400N HCI diluted to 300 ml volume (0.0200N) b. 25.0ml of 3.00M H2SO4 diluted to 100ml volume (0.750M) in 287ml of water and 100m final volume c. 8.00g of NaOH dissolved in 287ml of water and evaporated to 100 (2.00M)

Explanation / Answer

Ans 1 :

Molarity = no. of moles solute / volume of solution in Liter

So number of moles = volume in L x molarity

So here :

a) number of moles = 0.750 x 2.00

= 1.50 moles

Molar mass of HNO3 = 63.01 g/mol

So amount in grams = 1.50 x 63.01

= 94.5 g

b) number of moles = 0.450 x 1.25

= 0.562 moles

Molar mass of C12H22O11 = 342.3 g/mol

So amount in grams = 0.562 x 342.3

= 192 g

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