Question

calculate the number of sodium ions present in 1 kg of sodium carbonate

Answer 1

Moil of Na2CO3 required = 0.461*0.1017 = 0.04688 mol ( check your result 0.04588 mol) 
Mass of Na2CO3 required = 0.04688*105.99 = 4.9692g 

Calculate the amount (in mol) of sodium ions present in 95.0 mL of Solution A. 
1000mL of solution contains 0.04688 mol Na2CO3 
95.0mL solution contains 95.0/1000*0.04688 = 0.004454 mol Na2CO3 
Because 1 mol Na2CO3 produces 2 mol Na+ , 
0.004454 mol Na2CO3 will produce 2*0.004454 = 0.00891 mol Na+ ions